The QM I notes updates are strictly cosmetic (the book template is updated to that of classicthesis since it was originally posted). The chapters in QM II are reorganized a bit, grouping things by topic instead by lecture dates.

Motivation.

In class we were shown an adiabatic approximation where we started with (or worked our way towards) a representation of the form

where were normalized energy eigenkets for the (slowly) evolving Hamiltonian

In the problem sets we were shown a different adiabatic approximation, where are starting point is

For completeness, here’s a walk through of the general amplitude derivation that’s been used.

Guts

We operate with our energy identity once again

where

Bra’ing with , and split the sum into and parts

Again writing

We have

In this form we can make an “Adiabatic” approximation, dropping the terms, and integrate

or

Evaluating at , fixes the integration constant for

Observe that this is very close to the starting point of the adiabatic approximation we performed in class since we end up with

So, to perform the more detailed approximation, that started with 1.1, where we ended up with all the cross terms that had both and Berry phase dependence, we have only to generalize by replacing with .

Motivation

In the Fermi’s golden rule lecture we used the result for the integral of the squared function. Here is a reminder of the contours required to perform this integral.

Guts

We want to evaluate

We make a few change of variables

Now we pick a contour that is distorted to one side of the origin as in fig. 1.1

Fig 1.1: Contour distorted to one side of the double pole at the origin

We employ Jordan’s theorem (section 8.12 [1]) now to pick the contours for each of the integrals since we need to ensure the terms converges as for the part of the contour. We can write

The second two integrals both surround no poles, so we have only the first to deal with

Putting everything back together we have

On the cavalier choice of contours

The choice of which contours to pick above may seem pretty arbitrary, but they are for good reason. Suppose you picked for the first integral. On the big arc, then with a substitution we have

This clearly doesn’t have the zero convergence property that we desire. We need to pick the contour for the first (positive exponent) integral since in that range, is always negative. We can however, use the contour for the second (negative exponent) integral. Explicitly, again by example, using contour for the first integral, over that portion of the arc we have

Motivation.

that can be used to solve for a particular solution this differential equation via convolution

had the value

Let’s try to verify this.

Guts

Application of the Helmholtz differential operator on the presumed solution gives

When .

To proceed we’ll need to evaluate

Writing we start with the computation of

We see that we’ll have

Taking second derivatives with respect to we find

Our Laplacian is then

Now lets calculate the derivatives of . Working on again, we have

So we have

Taking second derivatives with respect to we find

So we find

or

Inserting this and into 2.8 we find

This shows us that provided we have

In the neighborhood of .

Having shown that we end up with zero everywhere that we are left to consider a neighborhood of the volume surrounding the point in our integral. Following the Coulomb treatment in section 2.2 of [1] we use a spherical volume element centered around of radius , and then convert a divergence to a surface area to evaluate the integral away from the problematic point

We make the change of variables . We add an explicit suffix to our Laplacian at the same time to remind us that it is taking derivatives with respect to the coordinates of , and not the coordinates of our integration variable . Assuming sufficient continuity and “well behavedness” of we’ll be able to pull it out of the integral, giving

Recalling the dependencies on the derivatives of in our previous gradient evaluations, we note that we have

so with , we can rewrite our Laplacian as

This gives us

To complete these evaluations, we can now employ a spherical coordinate change of variables. Let’s do the volume integral first. We have

To evaluate the surface integral we note that we’ll require only the radial portion of the gradient, so have

Our area element is , so we are left with

Putting everything back together we have

But this is just

This completes the desired verification of the Green’s function for the Helmholtz operator. Observe the perfect cancellation here, so the limit of can be independent of how large is made. You have to complete the integrals for both the Laplacian and the portions of the integrals and add them, before taking any limits, or else you’ll get into trouble (as I did in my first attempt).

Motivation.

I liked one of the adiabatic pertubation derivations that I did to review the material, and am recording it for reference.

Build up.

In time dependent pertubation we started after noting that our ket in the interaction picture, for a Hamiltonian , took the form

Here we have basically assumed that the time evolution can be factored into a portion dependent on only the static portion of the Hamiltonian, with some other operator , providing the remainder of the time evolution. From 2.1 that operator is found to behave according to

but for our purposes we just assumed it existed, and used this for motivation. With the assumption that the interaction picture kets can be written in terms of the basis kets for the system at we write our Schr\”{o}dinger ket as

where are the energy eigenkets for the initial time equation problem

Adiabatic case.

For the adiabatic problem, we assume the system is changing very slowly, as described by the instantanious energy eigenkets

Can we assume a similar representation to 2.3 above, but allow to vary in time? This doesn’t quite work since are no longer eigenkets of

Operating with does not give the proper time evolution of , and we will in general have a more complex functional dependence in our evolution operator for each . Instead of an dependence in this time evolution operator let’s assume we have some function to be determined, and can write our ket as

Operating on this with our energy operator equation we have

Here I’ve written . In our original time dependent pertubaton the term was , so this killed off the . If we assume this still kills off the , we must have

and are left with

Bra’ing with we have

or

The LHS is a perfect differential if we introduce an integration factor , so we can write

This suggests that we want to form a new function

or

Plugging this into our assumed representation we have a more concrete form

Writing

this becomes

A final pass.

Now that we have what appears to be a good representation for any given state if we wish to examine the time evolution, let’s start over, reapplying our instantaneous energy operator equality

Bra’ing with we find

Since the first and third terms cancel leaving us just

where and .

Summary

We assumed that a ket for the system has a representation in the form

where and are given or to be determined. Application of our energy operator identity provides us with an alternate representation that simplifes the results

Born approximation.

It turns out that finding the Green’s function is not so hard. Note the following, for , we have

(where a zero subscript is used to mark the case). We know this Green’s function from electrostatics, and conclude that

For we can easily show that

This is correct for all because it also gives the right limit as . This argument was first given by Lorentz. We can now write our particular solution

This is of no immediate help since we don’t know and that is embedded in .

Now look at this for

We get

where

If the scattering is weak we have the Born approximation

or

Should we wish to make a further approximation, we can take the wave function resulting from application of the Born approximation, and use that a second time. This gives us the “Born again” approximation of

Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

Scattering. Recap

READING: section 19, section 20 of the text [1].

We used a positive potential of the form of figure (\ref{fig:qmTwoL22:qmTwoL22fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL22fig1}
\caption{A bounded positive potential.}
\end{figure}

for

for

integrate these equations back to .

For

where both and are proportional to , dependent on .

There are cases where we can solve this analytically (one of these is on our problem set).

Alternatively, write as (so long as )

latex x x_2$}\end{array}\end{aligned} \hspace{\stretch{1}}(2.8)$

Now want to consider the problem of no potential in the interval of interest, and our window bounded potential as in figure (\ref{fig:qmTwoL22:qmTwoL22fig3})

Chapter 17.

\begin{itemize}
\item Page 297. (17.38). appears to be off by a factor of since .
\item Page 311. (17.134). missing on the term in the integral.
\item Page 311. (17.136). First term (non-integral part) should be negated.
\item Page 312. (17.144). Sign on before sum positive instead of negative.
\item Page 313. (17.149). missing.
\item Page 313. (17.152,17.154). extra bra around the bra.
\item Page 313. (17.153). bra missing on
\end{itemize}

Chapter 24.

\begin{itemize}
\item Page 450. (24.6). should be .
\item Page 452. (24.18). In the case should be instead of (although what's in the text is strictly still correct since it only changes the phase of the wavefunction).

\item Page 455. (24.40). RHD should be multiplied by .

\item Page 460. (24.71). should be .
\item Page 460. Third paragraph. should be .
\item Page 460. (24.76). The integral should be , not . This messes up some of the subsequent stuff, unless there is also another compensating error. Note that one can check this easily since the derivative of is .
\end{itemize}

Chapter 25.

Chapter 26.

\begin{itemize}
\item Page 486. (26.60). ought to have braces and read .
\item Page 487. (26.67). in the position should be .
\item Page 489. before (26.84). For rotations about the imaginary axis was probably meant to be the i’th axis.
\item Page 495. (26.149,26.150). Looks like ‘s are missing (esp. compared to 26.144-145).
\item Page 495. (26.150). off by . () as is.
\item Page 496. (26.154). An extra in the integral, in between and the .
\item Page 498. (26.175). should be in the first line.
\item Page 498. (26.178). An factor has been lost in either (26.178) or (26.179).
\item Page 499. (26.190). minor: should be .
\item Page 450. (26.192). minor: should be , and should be .
\end{itemize}

Chapter 27.

\begin{itemize}
\item Page 503. (27.8). minor: bold . probably meant to be .
\item Page 504. (27.20). Missing factor on LHS.
\item Page 507. before (27.53). minor: Velocity missing bold.
\item Page 510. before (27.78). minor: periods in the two kets should be commas.
\item Page 510. (27.80). and should be interchanged (if is the polar angle then for that rotation, and for the rotation in the plane). The ‘s here should also be dropped.
\item Page 511. (27.81). Same as 27.80.
\item Page 511. (27.83). ‘s should be dropped.
\item Page 514. (27.109). minor: dot instead of cdot.
\item Page 515. (27.117). Same error as in (26.149-150). ‘s missing, and wrong sign on .
\end{itemize}